Question: Solve for $x$ and $y$ using elimination. ${x+6y = 58}$ ${-x+5y = 30}$
Solution: We can eliminate $x$ by adding the equations together when the $x$ coefficients have opposite signs. Add the equations together. Notice that the terms $x$ and $-x$ cancel out. $11y = 88$ $\dfrac{11y}{{11}} = \dfrac{88}{{11}}$ ${y = 8}$ Now that you know ${y = 8}$ , plug it back into $\thinspace {x+6y = 58}\thinspace$ to find $x$ ${x + 6}{(8)}{= 58}$ $x+48 = 58$ $x+48{-48} = 58{-48}$ ${x = 10}$ You can also plug ${y = 8}$ into $\thinspace {-x+5y = 30}\thinspace$ and get the same answer for $x$ : ${-x + 5}{(8)}{= 30}$ ${x = 10}$